What is the derivative of 3x(18x^4+(13/(x+1))) ?

The derivative of 3x(18x^4+(13/(x+1))) is (3(90x^6+180x^5+90x^4+13))/((x+1)^2)

What is the first derivative of 3x(18x^4+(13/(x+1))) ?

The first derivative of 3x(18x^4+(13/(x+1))) is (3(90x^6+180x^5+90x^4+13))/((x+1)^2)

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\frac{d}{d x} (3 x (18 x^{4} + \frac{13}{x + 1}))

\frac{3 ( 90 x ^{6} + 180 x ^{5} + 90 x ^{4} + 13 )}{( x + 1 ) ^{2}}

Solution steps

\frac{d}{d x} (3 x (18 x^{4} + \frac{13}{x + 1}))

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 3 \frac{d}{d x} (x (18 x^{4} + \frac{13}{x + 1}))

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x, g = 18 x^{4} + \frac{13}{x + 1}

= 3 (\frac{d x}{d x} (18 x^{4} + \frac{13}{x + 1}) + \frac{d}{d x} (18 x^{4} + \frac{13}{x + 1}) x)

\frac{d x}{d x} = 1

\frac{d}{d x} (18 x^{4} + \frac{13}{x + 1}) = 72 x^{3} - \frac{13}{( x + 1 ) ^{2}}

= 3 (1 \cdot (18 x^{4} + \frac{13}{x + 1}) + (72 x^{3} - \frac{13}{( x + 1 ) ^{2}}) x)

Simplify

3 (1 \cdot (18 x^{4} + \frac{13}{x + 1}) + (72 x^{3} - \frac{13}{( x + 1 ) ^{2}}) x) : \frac{3 ( 90 x ^{6} + 180 x ^{5} + 90 x ^{4} + 13 )}{( x + 1 ) ^{2}}

= \frac{3 ( 90 x ^{6} + 180 x ^{5} + 90 x ^{4} + 13 )}{( x + 1 ) ^{2}}

\frac{d}{d x} (ln (sin (x^{3})))

x y = - 121

y = 12 e^{2} x

3 + 7 t + t^{2}

x \to - 8 + lim (\frac{- 9 x}{x + 8})

What is the derivative of 3x(18x^4+(13/(x+1))) ?
The derivative of 3x(18x^4+(13/(x+1))) is (3(90x^6+180x^5+90x^4+13))/((x+1)^2)
What is the first derivative of 3x(18x^4+(13/(x+1))) ?
The first derivative of 3x(18x^4+(13/(x+1))) is (3(90x^6+180x^5+90x^4+13))/((x+1)^2)